proof of Rouché’s theorem

Consider the integral

$$N(\lambda )=\frac{1}{2\pi i}{\oint }_C\frac{f^{\prime }(z)+\lambda g^{\prime }(z)}{f(z)+\lambda g(z)}𝑑z$$

where $0\le \lambda \le 1$. By the hypotheses, the function $f+\lambda g$ is non-singular on $C$ or on the interior of $C$ and has no zeros on $C$. Hence, by the argument principle, $N(\lambda )$ equals the number of zeros (counted with multiplicity) of $f+\lambda g$ contained inside $C$. Note that this means that $N(\lambda )$ must be an integer.

Since $C$ is compact, both $|f|$ and $|g|$ attain minima and maxima on $C$. Hence there exist positive real constants $a$ and $b$ such that

$$|f(z)|>a>b>|g(z)|$$

for all $z$ on $C$. By the triangle inequality, this implies that $|f(z)+\lambda g(z)|>a-b$ on $C$. Hence $1/(f+\lambda g)$ is a continuous function of $\lambda$ when $0\le \lambda \le 1$ and $z\in C$. Therefore, the integrand is a continuous function of $C$ and $\lambda$. Since $C$ is compact, it follows that $N(\lambda )$ is a continuous function of $\lambda$.

Now there is only one way for a continuous function of a real variable to assume only integer values – that function must be constant. In particular, this means that the number of zeros of $f+\lambda g$ inside $C$ is the same for all $\lambda$. Taking the extreme cases $\lambda =0$ and $\lambda =1$, this means that $f$ and $f+g$ have the same number of zeros inside $C$.